Angular momentum of a projectile

Today we look at an interesting problem that helps us feel closely the concept of angular momentum of a moving point. So let us dive in.

Now, we can definitely solve this using our sure-shot method of writing out the position vector r (with the relevant point: P, Q or R as the origin in each case) and the velocity vector v of the projectile as functions of time, and then performing the operation r x (mv) for the angular momentum L. Once we have L, we can analyse the function and comment on the things demanded in the problem.

But that method is not what we are here today for. Obviously.

If we cannot attack the problem in this basic way, what else do we have? Can we take the cross-product in some other, fancy way? Maybe, but I don't know how. Can we find the perpendicular distance of the particle's instantaneous velocity vector from the given origin (P, Q or R)? That seems quite difficult and lengthy, and rightly so.

One beautiful trick to simplify this problem is to not look directly at the angular momentum L but at a closely related quantity: the torque of gravity acting on the particle, because it is very easy to find. Since gravity is the only force acting, this torque is the rate at which L changes. 

The torque is easy to find because gravity acts vertically, so only the horizontal position of the particle (measured relative to P, Q or R) matters for the torque.

1. about P, the torque is initially zero, and then grows in the clockwise direction continuously until the projectile returns to ground. Now, since initially, the angular momentum about P (L_P ) is zero, it grows continuously in magnitude and is always directed clockwise until the particle returns to ground.
2. about Q, the torque acts anticlockwise for some time, becomes zero at the instant the particle is vertically above R, and then acts clockwise for the rest of the time. But it is easy to see that L_Q is clockwise throughout. So, its magnitude is decreased by the anticlockwise torque for some time, then increased by the clockwise torque.
3. about R, the torque of gravity always acts in the clockwise direction, similar to about P. But about R, initially the angular momentum is anti-clockwise, and just before the particle falls to the ground, it is clockwise. Therefore, the clockwise torque initially decreases the magnitude of L_R to zero, and then increases it in the clockwise direction. 

That can seem quite dense the first time you read it, so re-reading until you agree with everything said is a good idea.

Using the above analysis, we can draw rough sketches as shown below. Time in on the horizontal axis, and clockwise angular momentum is taken positive.

T here is of course the total time of flight. Why all three curves intersect at the same point and what the time T₀ is will be explained later in this post.

What is the shape of these curves? Since the torque varies linearly with the x-displacement, which varies linearly with time, the torque varies linearly with time. So, the angular momentum - time curves are parabolas.

Also, pay attention to the slope of these curves. What are they at t = 0? And how do they change with time?

Now for part (c) of the problem. Again, we have a nice way to cut through the mathematical detail and look at the problem such that it is solved in a couple of steps.

Here, r_P and r_Q are the position vectors of the moving particle with P and Q taken as origin respectively. 

Now, (r_P - r_Q) is constant. It is simply a vector with its head at Q and its tail at P (you might think it should be the other way around, but remember that r_P and r_Q are the position vectors of the moving particle with P and Q taken as origin, and NOT the position vectors of P and Q themselves). It is horizontal, and therefore our condition is met when the velocity of the particle is also horizontal. This is when the particle is at its highest point. The same analysis can be done for the pair Q and R, and for the pair P and R. The angular momentum about all three points are all equal to each other when the particle's velocity is horizontal. In hindsight this looks obvious, since at this instant, the perpendicular distance of the velocity vector from the point of reference is measured vertically, and the height of the particle is the same relative to P, Q or R.

So the time instant T₀ in the graph when the three curves all intersect is half of the total time of flight.

We have done some interesting things without doing any detailed maths at all, and it was fun (for you too, I hope ☺). You can consider them tricks or shortcuts, but they are just ways to use the basic fundamentals to make solving easier. 

We should also be able to write everything in detail. Try to write the angular momentum about P, Q and R as functions of time, as I have done below.

1. Torque is the rate of change of angular momentum, and hence the change in angular momentum is found from the time integral of torque:

2. Measuring the instantaneous horizontal separation of the particle from P, Q and R respectively (the separation from the point of projection P is x(t)), and then writing out the torques (clockwise positive).

3. Integrating to find the expressions for angular momentum.

We can deduce all the things asked for (and more) from these time functions: for example, the functions are all second degree, and so their graphs are parabolas, and so on.

So, we have seen both the clever and short method, and the long and trusted method. It is important to understand both, and practice both. Aside from other benefits, it can teach us how and where either or both of them can be used.

Cheers, and I hope you keep enjoying doing your physics!